Lemma 107.2.8. In Situation 107.2.1. Let $x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ be a morphism such that $\mathop{\mathrm{Spec}}(k) \to S$ is of finite type with image $s$. Let $A$ be a versal ring to $\mathcal{X}$ at $x_0$. If $\mathcal{O}_{S, s}$ is a G-ring, then we may find a smooth morphism $U \to \mathcal{X}$ whose source is a scheme and a point $u_0 \in U$ with residue field $k$, such that

1. $\mathop{\mathrm{Spec}}(k) \to U \to \mathcal{X}$ coincides with the given morphism $x_0$,

2. there is an isomorphism $\mathcal{O}_{U, u_0}^\wedge \cong A$.

Proof. Let $(\xi _ n, f_ n)$ be the versal formal object over $A$. By Artin's Axioms, Lemma 98.9.5 we know that $\xi = (A, \xi _ n, f_ n)$ is effective. By assumption $\mathcal{X}$ is locally of finite presentation over $S$ (use Morphisms of Stacks, Lemma 101.27.5), and hence limit preserving by Limits of Stacks, Proposition 102.3.8. Thus Artin approximation as in Artin's Axioms, Lemma 98.12.7 shows that we may find a morphism $U \to \mathcal{X}$ with source a finite type $S$-scheme, containing a point $u_0 \in U$ of residue field $k$ satisfying (1) and (2) such that $U \to \mathcal{X}$ is versal at $u_0$. By Lemma 107.2.6 after shrinking $U$ we may assume $U \to \mathcal{X}$ is smooth. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).